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From: phil@ux1.cso.uiuc.edu (Phil Howard KA9WGN)
Newsgroups: comp.unix.questions
Subject: Re: Half of a ls -l ?
Message-ID: <1991May12.232213.9435@ux1.cso.uiuc.edu>
Date: 12 May 91 23:22:13 GMT
References: <1991May12.085633.10066@nuchat.sccsi.com>
Organization: University of Illinois at Urbana
Lines: 36

robby@nuchat.sccsi.com (Robert Oliver Jr.) writes:

>  Has anybody out there ever printed out the size,date and name of a file
>without the permitions and owners just the size,date & name.  I would like 
>to know how you did it!  

Assuming you want time with the date, you can pipe "ls -l" through this
or some approximation of it to match your ls's output format:

----CUT-HERE--------CUT-HERE--------CUT-HERE--------CUT-HERE--------CUT-HERE----
#!/usr/local/bin/awk -f
{
    if( NF>7 ) {
        z=substr("         ",length($4)) $4;
        z=z substr("   ",length($5)) $5;
        z=z substr("  ",length($6)) $6;
        z=z substr("     ",length($7)) $7;
        z=z " " $8;
        print z;
    };
};
----CUT-HERE--------CUT-HERE--------CUT-HERE--------CUT-HERE--------CUT-HERE----

If you don't want the time, leave out the appropriate field.
If you new awk is somewhere else, change the first line.
Gawk may work, too, but I didn't try it.

Watch out for some versions of ls that, when faced with a very high link
count, will jam the link count number and the privileges together as one
single field.  To avoid that you might have to extract the files based on
a combination of column number and whitespace, rather than whitespace alone.
-- 
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/ Phil Howard -- KA9WGN -- phil@ux1.cso.uiuc.edu   |  Guns don't aim guns at  \
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